Inequality comes with a signs < (less than), > (greater than),

\(( \leq ) \; (less \; than \; equal \; to), \; (\geq ) \; (greater \; than \; equal \; to)\)

In GMAT, the questions will be completed and will be asked with modulus function.

**Steps to solve inequality question**

1. | Make an equation that corresponds to the inequality. |

2. | Solve this equation. |

3. | Fill in a number greater than and less than the solution then see which are correct with the inequality (you may use a number line to indicate these). |

4. | Write the solution of the inequality. |

When it comes to modulus the, the inequality will open in a different manner. This will be explained by following illustrations

If |x + 4| <9

Therefore, -9 < x +4 <9

If |x -5| >10

Therefore, x-5>10 or x-5<-10

Let’s look at some of the problem.

- Find the range of value of x for which.

\(\large x^{2} + 13x + 50 < 14\)

\(\large x^{2} + 13x + 36 < 0\)

\(\large (x + 4) (x + 9) < 0\)

After this form where the complete polynomial is expressed multiplication of single degree polynomial, we have to find the critical points. Critical points are the points where each single degree polynomials turn to zero.

So critical points are -4 and -9.

Hence we will check the behaviour around critical points. Please note that the sign of the polynomial remains constant for a region whether you take any number from that region.

In this case the complete number line is divided into 3 regions around critical points. I.e.

\(\large (-\infty , -9), [-9, -4) \; and \; [-4, \infty ).\)

From \((-\infty , -9)\) you can take -10. Hence the expression will be +ve

From \([-9, -4)\) you can take -6. Hence the expression will be -ve

From \( [-4, \infty ) \) you can take 0. Hence the expression will be +ve

Therefore the desired range is \([-9, -4)\). please do not forget to check the critical points. Here in this range, -9 is included on which the expression yield exactly 0. Hence it has to be excluded.

Therefore the answer will be \((-9, -4)\)

- Find the number of integers satisfying the equation

|x-5| + |x-10| < 30

Again due to critical points the number line will be divided in 3 regions. We will calculate the number of integers in each region

- Region 1: \((-\infty, 5]\)

The modulus will open like

-x + 5 – x +10 < 30

2x> -5

x > -2.5

Hence in this range x can take value like -2, -1, 0, 1, 2, 3, 4, ,5

- Region 2: (5, 10]

The modulus will open like

x – 5 -x+10 < 30

5<30

Hence all the value in this region will satisfy the equation. Therefore, x can take 6, 7, 8, 9, 10

- Region 3: \((10, -\infty]\)

The modulus will open like

X – 5 + x – 10 < 30

2x < 45

X < 22.5

Hence in this range x can take value like 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22

Therefore total number of integer that can be taken is 25.

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